Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 26

Answer

$$f'(t)=\frac{4-t^2}{2\sqrt{\frac{t}{t^2+4}}(t^2+4)^2}$$

Work Step by Step

$$f'(t)=\left(\sqrt{\frac{t}{t^2+4}}\right)'=\frac{1}{2\sqrt{\frac{t}{t^2+4}}}\left(\frac{t}{t^2+4}\right)'=\frac{1}{2\sqrt{\frac{t}{t^2+4}}}\frac{t'(t^2+4)-t(t^2+4)'}{(t^2+4)^2}=\frac{1}{2\sqrt{\frac{t}{t^2+4}}}\frac{1\cdot(t^2+4)-t\cdot2t}{(t^2+4)^2}=\frac{4-t^2}{2\sqrt{\frac{t}{t^2+4}}(t^2+4)^2}$$
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